FIG. 5-1 shows the simplified equivalent circuit for a coil in the presence of an ac magnetic field. The magnetic field induces voltage of magnitude, VB, in the coil. The coil has an inductance, LC, a resistance, RC, and a capacitance, CC. For simplicity consider a coil of circular cross section in a uniform ac magnetic field parallel to its axis whose amplitude is given by Bsin(ωt). The coil resistance is given by:RC=2πrρN  (5-1)Where r is the radius of the coil, N is the number of turns in the coil, r is the radius of the coil, and ρ is the wire resistance per unit length. The coil inductance is given by:LC=μ0N2πr2K/1  (5-2)Where μ0 is the permeability of free space, 1 is the coil length, and K is an inductive shape constant. K is generally in the range of 0.1 to 1.0 for real coils. There is no simple formula for the coil capacitance but it generally proportional to NX, where x is between 1 and 2. The magnitude of the voltage induced in the coil is given by:VB=−Nπr2ωB  (5-3)The magnitude of the open circuit voltage across the coil, VC, is given by:VC=VB/[(1−ω2LCCC)+jωRCCC]  (5-4)As ω goes to zero VC goes to VB which goes to zero. Also as ω goes to infinity VC goes to zero. This limits the range of usefulness. In addition when ω is equal to (1/LCCC)1/2, then Vc is given by:VC=−Nπr2ωB/(jωRCCC)  (5-5)This represents a resonance. Since the coil resistance and capacitance is small the Q of this resonance is high. The useful range of operation is thus limited to well below the resonance. If the number of turns and or radius of the coil are increased to improve the sensitivity then the resonant frequency is moved lower further reducing the useful frequency band of the sensor.
If the coil is used instead as a current source, the short circuit coil current, Is, is given by:IS=−Nπr2ωB/(RC+jωLC)  (5-6)This is a highpass response with a corner frequency, ω0, given by, RC/LC or:ω0=2ρl/(μ0NrK)  (5-7)As the frequency goes to zero the short circuit coil current goes to zero; however above the corner frequency, the current is independent of the frequency and proportional to the magnetic field magnitude. For the frequency above the corner frequency the short circuit coil current is given by:IS=jB1(μ0NK)  (5-8)The current mode provides a wide frequency band of operation above the corner frequency. In addition the corner of the band can be made as small as desired by decreasing the wire resistance, ρ, or by increasing the coil radius, r. This can be accomplished without changing the sensitivity of the coil's response, IS. It is also important to note that the coil capacitance, which was responsible for the resonance in the voltage mode case, has no affect on the current mode transfer function.